Problem: Complete the square to solve for $x$. $x^{2}-8x+16 = 0$
Answer: The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $-8$ , half of it is $-4$ , and squaring it gives us ${16}$ , our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x - 4 )^2 = 0$ Take the square root of both sides. $x - 4 = 0$ Isolate $x$ to find the solution(s). The solution is: $x = 4$ We already found the completed square: $( x - 4 )^2 = 0$